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I'm interested in functional equation. I've been thinking about the following functional equation: $$f(x^2+1)={f(x)}^2+1\ \ \ \cdots(\star).$$

I found several functions such as $f(x)=x, x^2+1, (x^2+1)^2+1,\cdots$. Then, I got interested the following question:

Find every natural number $n$ such that there exists a $n$-th polynomial which satisfies $(\star).$

Then, I got the following fact:

Fact: There exists a $2^k$-th polynomial which satisfies $(\star)$ for any non-negative integer $k$.

Proof: Let us define $\{f_k(x)\}$ as the following: $$f_0(x)=x, f_{k+1}(x)={f_k(x)}^2+1\ \ \ \cdots(\star\star).$$

Supposing that both $f_0(x)$ and $f_k(x)$ satisfy $(\star)$, then $f_{k+1}(x)$ also satisfies $(\star)$ because $$f_{k+1}(x^2+1)=\{f_k(x^2+1)\}^2+1=\{{f_k(x)}^2+1\}^2+1={f_{k+1}(x)}^2+1.$$

Hence, $\{f_k(x)\}$ satisfy $(\star)$. Since we know that the degree of $f_k(x)$ is $2^k$ inductively from $(\star\star)$, now the proof is completed.

However, I can't prove that there is no other $n$ such that there exists a $n$-th polynomial which satisfies $(\star)$.

Then, here is my question.

Question: Could you show me how to solve the above question?

mathlove
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    do you mean $n$-th degree polynomial? – obataku Sep 05 '13 at 12:28
  • Naively you might consider the case of 3rd order polynomials. If you were to substitute an arbitrary 3rd order polynomial $p_3(x)= ax^3+bx^2+cx+d$ into $(\star)$ we find $$a(x^2+1)^3+b(x^2+1)^2+c(x^2+1)+d = (ax^3+bx^2+cx+d)^2+1.$$ Expanded this is $$a\left(a-1\right) x^6 -2 a b x^5 +\left(3 a+b-b^2-2 a c\right) x^4 -2( b c + a d) x^3 +\left(3 a+2 b+c-c^2-2 b d\right) x^2 -2 c d x + a+b+c+d-d^2-1=0.$$ Is you're question equivalent to showing polynomials of this form (for each $n\neq2^k$) have no real solutions? – Brulboy Sep 05 '13 at 13:43
  • @Brulboy It would be, but already for $n=3$ trying to consider the polynomials would be a mess. – Andrew D Sep 05 '13 at 13:59
  • @oldrinb: Yes. I edited it. Thank you. – mathlove Sep 05 '13 at 14:26
  • @Brulboy: I think yes. – mathlove Sep 05 '13 at 14:29
  • @mathlove : dinoboy's or miracle173's answers there show that there is one polynomial (the obvious one) of degree $n = 2^k$ for each $k \ge 0$ ; and no other. – mercio Sep 05 '13 at 14:39
  • @merico: Oh! You're right. Their answers include what i want. Thank you. – mathlove Sep 05 '13 at 14:52
  • Dear @mathlove : Yeah, the solutions at the other question are doing that. Check them out :) – rschwieb Sep 05 '13 at 14:55

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